Gravitational Field

GRAVITATIONAL FIELD

GRAVITATIONAL FORCE BETWEEN TWO MASSES (NEWTON’S LAW OF UNIVERSAL GRAVITATION)

A gravitational field is a region of space surrounding a body that has the property of mass. Sir Isaac Newton, in 1666, propounded the universal law of gravitation.

The law states that, ”the force of attraction between two given particles of masses M and m is directly proportional to the product of the masses and inversely proportional to the square of their distance of separation”.

This is the relationship between gravitational constant G and acceleration due to gravity g.

Therefore, we define gravitational field strength ‘g’ as force per unit mass. It is a vector quantity.

GRAVITATIONAL POTENTIAL Vg

The gravitational potential Vg at a point is the work done in taking a unit mass from infinity to that point on the surface of the earth.

\(V_g = gr \) ——-(4) \(V_g = \frac{GM}{r^2} \times r \)

∴ \( V_g = \frac{GM}{r} \) ——-(5)

Where M is the mass of the earth and r is the radius of the earth of value 6.4 × 106m or 6400km

At any point, distance r from the centre of the earth, the gravitational potential experienced by a body of mass m is given by:

\(∴ V_g = −\frac{Gm}{r} \) ——-(6)

Since the potential at infinity is taken to be zero. (i.e, ∆potential \(= 0 -V_g = -V_g = -\frac{Gm}{r} \)

The negative sign indicates that potential at infinity is higher than the potential close to the mass, that is, Vg decreases as r increases.

ESCAPE VELOCITY V0

Consider a rocket of mass m placed at the centre of the earth’s surface O. If it is fired from that point so that it just escapes the earth’s gravitational field, it has a kinetic energy, k.e given as:

\(k.e = \frac{1}{2}mV_0^2 \) ——-(7)

But work is done(WD) in taking this rocket to a distance R so great that the gravitational field is negligibly weak.

∴ Work done \(= mg × distance\)

This work done must be equal to the kinetic energy of the rocket at the point of take off

Work done \(= \frac{1}{2}mV_0^2 \)

∴ Work done \(= mg × R = mgR\)

but gravitational intensity \(g = \frac{GM}{R^2} \)

Work done \(= m \Big (\frac{GM}{R^2} \Big ) R\)

Work done \(= \frac{GMm}{R}\)

But k.e = WD

\(∴\frac{1}{2}mV_0^2 = \frac{GMm}{R} \\ ∴ V_0^2 = \frac{2GM}{R}\)

∴ \(V_0 = \sqrt{\frac{2GM}{R}}\) ——-(8)

But \(gR = \frac{GM}{R}\)

Hence, \(V_0 = \sqrt{2gR}\) ——-(8)

We thus define the escape velocity V0 as the velocity which is sufficient enough for a body to just escape the earth’s gravitational field.

SATELLITES AND PARKING ORBIT

Satellites are bodies, natural or artificial which move in orbits around the moon or planets.Artificial satellites are made by man. Consider a satellite of mass m moving round the earth of mass M in an orbit as shown below:

If R is the radius of the earth, r is the radius of the orbit and v is the velocity with which the satellite is moving,

We have that centripetal force (due to the satellite) = gravitational force (due to the earth)

Since R = 6400km and g = 9.8m/s2, therefore, T = 5077.58s = 84.6mins. This is the period for an artificial satellite in the orbit above the earth.

If the period of the satellite in its orbit is equal to the period of the earth about its axis, which is 24hours, the satellite will stay at the same place above the earth as the earth rotates. Such orbit is called ‘’Parking Orbit’’.